Take Quadratic And Cubic, Get Sum And Product (Posted on 2007-08-24)

	Submitted by K Sengupta
Rating: 3.8000 (5 votes)
Solution:	(Hide)
From the second equation, we observe that x> y, z. Accordingly, from the first equation, we obtain x^2 < 6x, so that : x< 6. The RHS of the first equation is even, so that x = 2, 4. Suppose x=2. If so, we must have y+z = 0 in terms of the first equation.. This does not admit of any positive integral solution which is a contradiction. Suppose x =4. Then the first equation yields y+z = 4. Imposing the restriction y>=z, this has the solutions (y, z) = (3, 1); (2, 2) Thus, (x, y, z) = (4, 3, 1); (4,2, 2) correspond to all possible solution to the given problem. -------------------------------------------------------------------------- Praneeth provides a solution to the general case of determining all possible integer triplets that satisfy conditions of the problem in this location.

	Subject	Author	Date
	Alternative approach	FrankM	2008-01-19 21:51:13
	re(2): Solution (little thing to add)	Dej Mar	2007-08-25 09:11:05
	re: Solution (little thing to add)	Daniel	2007-08-25 02:15:58
	Solution	Praneeth	2007-08-24 15:23:57